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If the cube roots of unity are 1, ω, ω^{2}, then the roots of the equation (x – 1)^{3} + 8 = 0 are (1979)
(x – 1)^{3 } + 8 = 0
⇒ (x – 1)^{3} = – 8 = (– 2)^{3}
⇒ x – 1 = – 2
or 2ω or 2ω^{2}
⇒ x = 1, 1  2ω, 1 2ω^{2}
The smallest positive integer n for which (1980)
= 1 is
Now i^{n} = 1 ⇒ the smallest positive integral value of n should be 4.
The complex numbers z = x+ iy which satisfy the equation lie on....... (1981  2 Marks)
ATQ  x + iy – 5i  =  x + iy + 5i 
⇒  x + (y – 5) i  =  x + (y + 5) i 
⇒ x^{2 }+ (y – 5)^{2 }= x^{2} + (y + 5)^{2}
⇒ x^{2} + y^{2 }– 10y + 25 = x^{2} + y^{2} + 10y + 25
⇒ 20y = 0 ⇒ y = 0
∴ ‘a’ is the correct alternative.
If then (1982  2 Marks)
⇒ Re(z) < 0 and Im(z) = 0
∴ (b) is the correct choice.
The inequality z – 4 < z –2 represents the region given by (1982  2 Marks)
 z – 4  <  z – 2 
⇒  (x – 4) + iy  <  (x – 2) + iy 
⇒ (x – 4)^{2} + y^{2} < (x – 2)^{2} + y^{2}
⇒ – 8x + 16 < – 4x + 4
⇒ 4x – 12 > 0 ⇒ x > 3 ⇒ Re (z) > 3
If z = x + iy and ω = (1  iz) /( z  i) , then ω= 1 implies that, in the complex plane, (1983  1 Mark)
⇒ 1– iz  =  z – i  ⇒  1 – i (x + iy)  =  x + iy – i 
⇒  (y + 1) – ix  =  x + i (y – 1) 
⇒ x^{2} + (y + 1)^{2 }= x^{2} + (y – 1)^{2 }
⇒ 4y = 0 ⇒ y = 0 ⇒ z lies on real axis
The points z_{1}, z_{2}, z_{3} z_{4} in the complex plane are the vertices of a parallelogram taken in order if and only if (1983  1 Mark)
If vertices of a parallelogram are z_{1}, z_{2}, z_{3}, z_{4} then as diagonals bisect each other
= z_{1}+ z_{3} = z_{2}+ z_{4}
If a, b, c and u, v, w are complex numbers representing the vertices of two triangles such that c = (1 – r) a + rb and w = (1 – r)u + rv, where r is a complex number, then the two trian gles (1985  2 Marks)
Let ABC be the Δ with vertices a, b, c and PQR be the Δ with vertices u, v, w.
Then c = (1– r) a + rb.
⇒ c – a = r(b – a) ....(1)
⇒ w = (1– r) u + rv ....(2)
From (1) and (2) and
⇒ ΔABC ~ΔPQR.
If ω (≠1) is a cube root of unity and (1 + ω)^{7} = A + B ω then A and B are respectively (1995S)
(1 + ω)^{7} = A +Bω
⇒ ( ω^{2})^{7} = A +Bω (∵1 + ω +ω^{2} = 0)
⇒  ω^{14} = A +Bω ⇒  ω^{2} = A +Bω (∵ ω^{3}= 1)
⇒ 1 + ω = A +Bω ⇒ A = 1, B = 1
Let z and ω be two non zero complex numbers such that z = ω and Arg z+ Arggω = π, then z equals (1995S)
Q  z  =  ω  and argz = π  argω
Let
Let z and ω be two complex numbers such that z ≤1, ω ≤ 1 and z + iω =  z – i  = 2 then z equals (1995S)
Given that
⇒ z lies on perpendicular bisector of the line segment joining which is real ax is , being mirror images of each other..
∴ Im(z) = 0.
If z = x then
∴ (c) is the correct option.
For positive integers n_{1}, n_{2} the value of the expression (1 + i)^{n1} + (1 + i^{3})^{n1} + (1+ i^{5})^{n2} + (1+i^{7})^{n2} , where i = – is a real number if and only if (1996  1 Marks)
(1 + i)^{ n}1 + (1+ i3 )^{n}1 + (1+ i5 )^{n}2 + (1+i7 )^{n}2
= (1 + i) ^{n}1 + (1 – i)^{n}1 + (1 + i)^{n}2+ (1 –i )^{n}2
Using
and
We get the given expression as
= real number irrespective the values of n_{1} and n_{2}
∴ (d) is the most appropriate answer.
If i = , then 4 + 5 + 3 is equal to (1999  2 Marks)
= 1+2w + 3(1+ w + w^{2}) =
If arg(z) < 0, then arg (z)  arg(z) = (2000S)
arg (z) < 0 (given) ⇒ arg (z) = – θ
Now
z = r cos(θ) + i sin (θ) = r [cos (θ) i sin (θ)]
Again  z = r [cos (θ) i sin (θ)] = r [cos (πθ) +i sin (πθ)]
∴ arg ( z) = π θ;
Thus arg ( z)  arg(z) =π  θ  (θ) =π  θ +θ =π
If z_{1}, z_{2} and z_{3} are complex numbers such that (2000S)
z_{1} = z_{2} = z_{3} = = 1, then z_{1} +z_{2} +z_{3} is
 z_{1}  =  z_{2}  =  z_{3} = 1 (given)
Now,
Similarly
Now,
NOTE THIS STEP
Let z_{1 }an d z_{2} be n^{th }roots of unity which subtend a right angle at the origin . Then n must be of the form (2001S)
Let z = (1)^{1/ n} = (cos ^{2}kπ + i sin ^{2}kπ)^{1/n}
k, 0, 1, 2, .....,n 1.
Let
and
be the two values of z. s.t. they subtend ∠ of 90° at origin.
As k_{1} and k_{2} are integers and k_{1} ≠ k_{2.}
∴ n = 4k, k ∈ I
The complex numbers z_{1}, z_{2} and z_{3} satisfying are the vertices of a triangle which is
⇒ arg (cos( π /3) + i sin (π /3))
⇒ angle between z_{1} – z_{3} and z_{2}– z_{3} is 60°.
and
NOTE THIS STEP
⇒ The Δ with vertices z_{1}, z_{2 }and z_{3} is isosceles with vertical ∠60° . Hence rest of the two angles should also be 60° each.
⇒ Req. Δ is an equilateral Δ.
For all complex numbers z_{1}, z_{2} satisfying z_{1}=12 and  z_{2}34i = 5, the minimum value of z_{1}z_{2} is (2002S)
 z_{1} =12 ⇒ z_{1} lies on a circle with centre (0, 0) and radius 12 units, and  z_{2} – 3 – 4i  = 5
⇒ z_{2} lies on a circle with centre (3, 4) and radius 5 units.
From fig. it is clear that  z_{1}– z_{2}  i.e., distance between z_{1} and z_{2} will be min when they lie at A and B resp. i.e., O,C, B, A are collinear as shown.
Then z_{1}– z_{2} = AB = OA – OB = 12 – 2(5) = 2.
As above is the min, value, we must have  z_{1}– z_{2} ≥ 2.
If z = 1 and ω = ( where Z ≠ 1) , then Re(ω) is
Given that  z  = 1 and
Now we know that
(for  z  = 1)
⇒ Re (ω)=0
If ω(≠1) be a cube root of unity and (1 + ω^{2})^{n} = (1 + ω^{4})^{n}, then the least positive value of n is (2004S)
(1 +ω^{2})^{n} = (1 +ω^{4})^{n}
⇒ (ω)^{n} = (1+ ω)^{n} = (ω^{2})^{n} ⇒ ω^{n }= 1 ⇒ n =3
The locus of z which lies in shaded region (excluding the boundaries) is best represented by (2005S)
Here we observe that.
AB = AC = AD = 2
∴ BCD is an arc of a circle with centre at A and radius 2.
Shaded region is outer (exterior) part of this sector ABCDA.
∴ For any pt. z on arc BCD we should have  z – (– 1)  = 2
and for shaded region,  z + 1 > 2 ....(i)
For shaded region we also have p /4 < arg ( z + 1) <p /4
or  arg (z + 1)  <p /4 ...(ii)
Combining (i) and (ii), (a) is the correct option
a, b, c are integers, not all simultaneously equal and ω is cube root of unity (ω ≠ 1), then minimum value of a + bω + cω^{2} is (2005S)
Given that a, b, c are integers not all equal, w is cube root of unity ≠ 1, then
 a + bω +cω^{2}
∴ The min value is obtained when any two are zero and third is a minimum magnitude integer i.e. 1.
Thus b = c = 0, a = 1 gives us the minimum value 1
Let ω = +i then the value of the det.
(2002  2 Marks)
Operating R_{1} + R_{2} + R_{3}, we get
If is purely real where w = α + iβ, β ≠ 0 and z ≠ 1,then the set of the values of z is (2006  3M, –1)
is purely real
⇒  z ^{2} = 1 (∵ ω = α + iβ and β ≠ 0)
⇒  z  = 1 also given z ≠ 1
∴ The required set is {z :  z  =1, z ≠ 1}
= 3ω (ω 1)
A man walks a distance of 3 units from the origin towards the northeast (N 45° E) direction. From there, he walks a distance of 4 units towards the northwest (N 45° W) direction to reach a point P. Then the position of P in the Argand plane is (2007 3 marks)
If z = 1 and z ≠ ± 1, then all the values of lie on (2007 3 marks)
Given  z  = 1 and z ≠ ± 1
To find locus of
We have
purely imaginary number
∴ ω must lie on y – axis.
A particle P starts from the point z_{0} = 1 + 2i, where i = .It moves horizontally away from origin by 5 units and then vertically away from origin by 3 units to reach a point z1.From z1 the particle moves units in the direction of the vector and then it moves through an angle in anticlockwise direction on a circle with centre at origin, to reach a point z_{2}. The point z_{2} is given by (2008)
The initial position of point is Z_{0 }= 1+ 2i
∴ Z_{1} = (1 + 5) + (2 + 3) i = 6 + 5i
Now Z_{1} is moved through a distance of 2 units in the direction
∴ It becomes
Now OZ_{1}' is rotated through an angle inanticlockwise direction, therefore Z_{2} = iZ_{1}' =  6+7i
Let z = cosθ + i sinθ. Then the value of
at θ = 2° is (2009)
z = cosθ + i sinθ
= cos(2m  1)θ + i sin(2m  1)θ
= sinθ + sin 3θ + sin 5θ + ......+ upto15 terms
Let z = x + iy be a complex number where x and y are integers.Then the area of the rectangle whose vertices are the roots of the equation : (2009)
Given z = x + iy where x and y are integer
Also
⇒ ( x^{2} +y^{2}) ( x^{2} y^{2})= 175
⇒ ( x^{2} +y^{2}) ( x^{2} y^{2}) = 25x7 ...(i)
or(x^{2} + y^{2}) (x^{2} – y^{2}) = 35 × 5 ...(ii)
∵ x and y are integers,
∴ x^{2} +y^{2}= 25 and x^{2}  y^{2 }= 7 [From eq (i)]
⇒ x^{2 }= 16 and y 2 = 9 ⇒ x = ± 4 and y =±3
∴ Vertices of rectangle are (4, 3) , (4, – 3), (– 4, – 3) , (– 4, 3).
So, area of rectangle = 8 × 6 = 48 sq. units Now from eq. (ii)
or x^{2} + y^{2} = 35 and x^{2} – y^{2} = 5
⇒ x^{2} = 20, which is not possible for any integral value of x
Let z be a complex number such that the imaginary part of z is nonzero and a = z^{2} + z + 1 is real. Then a cannot take the value
(2012)
∵ Im (z) ≠ 0
⇒ z is non real and equation z^{2} + z + (1 a )=0 will have non real roots, if D < 0
⇒ 1 – 4(1 – a) < 0 ⇒ 4a < 3 ⇒ a <
∴ a can not take the value
Let complex numbers a and lie on circles (x – x_{0})^{2} + (y – y_{0})^{2} = r^{2} and (x – x_{0})^{2 }+ (y – y_{0})^{2} = 4r^{2}. respectively. If z_{0 }= x_{0} + iy_{0 }satisfies the equation
2 z_{0}^{2} = r^{2} + 2, then α = (JEE Adv. 2013)
As α lies on the circle (x – x_{0})^{2} + (y – y_{0})^{2} = r^{2}
∴ α – z_{0}^{2} = r^{2}
(i)
Also lies on the circle (x – x_{0})^{2 }+ (y – y_{0})^{2} = 4r^{2}
Subtracting eqn (i) from (ii) we get
or
Using we get
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